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          数项级数审敛问题
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              <time title="Created: 2024-10-26 01:00:03" itemprop="dateCreated datePublished" datetime="2024-10-26T01:00:03+00:00">2024-10-26</time>
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        <h2 id="题目">题目</h2>
<p>已知数项级数 <span class="math inline">\(\sum\limits_{n=0}^{\infty}
a_{n}\)</span> 收敛，则下列级数 <span class="math display">\[\sum_{n=0}^{\infty}(-1)^{n} a_{n},\quad
\sum_{n=0}^{\infty}\left(a_{n}\right)^{2},\quad\sum_{n=0}^{\infty}\sin
\left(a_{n}\right),\quad\sum_{n=0}^{\infty}\ln\left(1+a_{n}\right)\]</span>
中收敛级数的数量是（ ） <span class="math display">\[\text{A}.1\quad\text{B}.2\quad\text{C}.3\quad\text{D}.4\]</span></p>
<ul>
<li>答案选择 A 项</li>
</ul>
<blockquote>
<p>该试题来自<strong>中国科学院大学 2019 年 601高等数学(甲)</strong>
选择第7题。可以从<a target="_blank" rel="noopener" href="https://gitee.com/ylxdxx/AM601-kaoyan">中科院601高等数学甲考研真题集(gitee.com)</a>仓库获得</p>
</blockquote>
<h2 id="题解">题解</h2>
<ul>
<li>对于 <span class="math inline">\({\displaystyle\sum_{n=0}^{\infty}(-1)^{n}
a_{n}}\)</span> 可以举出 <span class="math inline">\(a_n=(-1)^{n}\cfrac{1}{n+1}\)</span> 使得发散</li>
<li>对于 <span class="math inline">\({\displaystyle\sum_{n=0}^{\infty}(a_{n})^2}\)</span>
可以举出 <span class="math inline">\(a_n=(-1)^{n}\cfrac{1}{\sqrt{n+1}}\)</span>
使得发散</li>
<li>对于 <span class="math inline">\({\displaystyle\sum_{n=0}^{\infty}
\ln \left(1+a_{n}\right)}\)</span> 可以泰勒展开为 <span class="math inline">\({\displaystyle\sum_{n=0}^{\infty}\left(a_n-\cfrac{1}{2}a_n^2+o(a_n^2)\right)}\)</span>，注意到
<span class="math inline">\((a_{n})^2\)</span> 同理可以发散</li>
<li><strong>关键在于</strong> <span class="math inline">\({\displaystyle\sum_{n=0}^{\infty} \sin
\left(a_{n}\right)}\)</span> <strong>是否也可以发散</strong></li>
</ul>
<p>构造 <span class="math inline">\({a_n=\cfrac{1}{\sqrt[3]{n}}\left(\cfrac{2\pi}{3}n\right)}\)</span>
使得 <span class="math inline">\({\displaystyle\sum_{n=1}^{\infty} \sin
\left(a_{n}\right)}\)</span> 发散，大致思路如下：</p>
<ol type="1">
<li>证明 <span class="math inline">\(\sum\limits_{n=1}^{\infty}
a_{n}\)</span> 收敛
<ol type="1">
<li><span class="math inline">\(\alpha_n=\cfrac{1}{\sqrt[3]{n}}\)</span>
递减，且 <span class="math inline">\({\displaystyle\lim_{n\to\infty}
a_n=0}\)</span><br>
</li>
<li><span class="math inline">\(\beta_n=\cos\left(\cfrac{2\pi}{3}n\right)\)</span>，有
<span class="math inline">\({\displaystyle\left|\sum_{n=1}^{\infty}\beta_n\right|\leq1}\)</span></li>
<li>由狄利克雷判别法得到 <span class="math inline">\(\sum\limits_{n=1}^{\infty}
\alpha_{n}\beta_{n}\)</span> 收敛</li>
</ol></li>
<li>证明 <span class="math inline">\({\displaystyle\sum_{n=1}^{\infty}
\sin \left(a_{n}\right)}\)</span> 发散
<ol type="1">
<li>泰勒展开为 <span class="math inline">\({\displaystyle\sum_{n=1}^{\infty}\left(a_n-\cfrac{1}{6}a_n^3+o(a_n^3)\right)}\)</span>
，关注 <span class="math inline">\({\displaystyle\sum_{n=1}^{\infty}a_{n}^3}\)</span>
的敛散性 <span class="math display">\[{a_n^3=\cfrac{1}{n}
{\cos}^3\left(\cfrac{2\pi}{3}n\right)}=\left\{
\begin{array}{**cl**}
-\cfrac{1}{8n}&amp;&amp;n=3k-2,3k-1\\
\cfrac{1}{n}&amp;&amp;n=3k\\
\end{array}
\right.\quad k=1,2,3,\dots\]</span></li>
<li>逐三项求和 <span class="math display">\[\sum_{n=1}^{\infty}a_{n}^3=\sum_{k=1}^{\infty}-\cfrac{1}{8(3k-2)}-\cfrac{1}{8(3k-1)}+\cfrac{1}{3k}=\sum_{k=1}^{\infty}\cfrac{54k^2-63k+16}{8(3k-2)(3k-1)3k}\]</span></li>
<li>比值审敛法得 <span class="math inline">\({\displaystyle\lim_{n\to\infty}na^3_n}=\cfrac{1}{4}\)</span>
，所以 <span class="math inline">\({\displaystyle\sum_{n=1}^{\infty}a_{n}^3}\)</span>
发散，得 <span class="math inline">\({\displaystyle\sum_{n=1}^{\infty}
\sin \left(a_{n}\right)}\)</span> 发散</li>
</ol></li>
</ol>
<h2 id="思考">思考</h2>
<ul>
<li>数项级数 <span class="math inline">\(\sum\limits_{n=0}^{\infty}
a_{n}\)</span> 收敛一般不能轻易推出 <span class="math inline">\(\sum\limits_{n=0}^{\infty} f(a_{n})\)</span>
收敛</li>
<li>对于正项级数，我们有不少审敛方法，而对于更加一般的数项级数，在高数中只有对交错项级数给出了莱布尼兹判别法。考研竞赛中若遇到数项级数审敛的题目，由于我们缺乏处理数项级数的工具，应该考虑举出反例。利用泰勒展开提取高次项，参考狄利克雷判别法构造收敛函数。</li>
</ul>

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